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geometry hon - garcia (lms) 2 - 4: teacher_s_slope_parallel lines_p8 (l…

Question

geometry hon - garcia (lms)
2 - 4: teacher_s_slope_parallel lines_p8 (lms graded)
slope of p = - 3
slope of q =
slope of r = 1.36
slope of m = 0
slope of n =

Explanation:

Step1: Recall slope - formula

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Find slope of line $q$

Let's assume two points on line $q$ are $(0.4,-3)$ and $(4,-12)$. Then $x_1 = 0.4,y_1=-3,x_2 = 4,y_2=-12$. Substitute into the slope formula: $m_q=\frac{-12-(-3)}{4 - 0.4}=\frac{-12 + 3}{3.6}=\frac{-9}{3.6}=-2.5$.

Step3: Find slope of line $n$

Let's assume two points on line $n$ are $(-5,-6.8)$ and $(0.4,-3)$. Then $x_1=-5,y_1 = - 6.8,x_2=0.4,y_2=-3$. Substitute into the slope formula: $m_n=\frac{-3-(-6.8)}{0.4-(-5)}=\frac{-3 + 6.8}{0.4 + 5}=\frac{3.8}{5.4}=\frac{19}{27}\approx0.704$.

Answer:

slope of $q=-2.5$
slope of $n=\frac{19}{27}$